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Solve the following undefined questions. a. 2y / (6x + 4)(3x - 9) b. 20 / (x² + 5x + 6)

Mathematics

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Step-by-step solution

12 steps

Part 2· 3 steps

Solve the right triangle using trigonometry (adjacent side = 5 cm, θ=20∘\theta = 20^\circ, opposite = xx cm, hypotenuse = yy cm)

  1. Use tangent for opposite side.

    tan⁡θ=x5\tan \theta = \frac{x}{5} tan⁡20∘=x5\tan 20^\circ = \frac{x}{5} x=5tan⁡20∘≈5×0.3640=1.82 cmx = 5 \tan 20^\circ \approx 5 \times 0.3640 = 1.82\ \mathrm{cm}

  2. Use cosine for hypotenuse.

    cos⁡θ=5y\cos \theta = \frac{5}{y} y=5cos⁡20∘y = \frac{5}{\cos 20^\circ} y≈50.9397≈5.32 cmy \approx \frac{5}{0.9397} \approx 5.32\ \mathrm{cm}

  3. Verify with sine.

    sin⁡20∘=xy≈1.825.32≈0.3420\sin 20^\circ = \frac{x}{y} \approx \frac{1.82}{5.32} \approx 0.3420

    Matches sin⁡20∘\sin 20^\circ.

Part 3· 3 steps

Solve the triangle using law of cosines (sides a=6a=6 cm, b=10b=10 cm, ∠C=10∘\angle C=10^\circ, find side cc, then angles)

  1. Law of cosines for side cc.

    c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C c2=62+102−2⋅6⋅10⋅cos⁡10∘c^2 = 6^2 + 10^2 - 2 \cdot 6 \cdot 10 \cdot \cos 10^\circ c2=36+100−120⋅0.9848≈136−118.18=17.82c^2 = 36 + 100 - 120 \cdot 0.9848 \approx 136 - 118.18 = 17.82 c≈17.82≈4.22 cmc \approx \sqrt{17.82} \approx 4.22\ \mathrm{cm}

  2. Law of cosines for ∠A\angle A.

    cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc} cos⁡A=102+4.222−622⋅10⋅4.22≈100+17.81−3684.4≈81.8184.4≈0.9695\cos A = \frac{10^2 + 4.22^2 - 6^2}{2 \cdot 10 \cdot 4.22} \approx \frac{100 + 17.81 - 36}{84.4} \approx \frac{81.81}{84.4} \approx 0.9695 ∠A≈cos⁡−1(0.9695)≈14.1∘\angle A \approx \cos^{-1}(0.9695) \approx 14.1^\circ

  3. Find ∠B=180∘−10∘−14.1∘=155.9∘\angle B = 180^\circ - 10^\circ - 14.1^\circ = 155.9^\circ.

Part 4· 2 steps

Solve x=y+5sx = \frac{y + 5}{s} for ss in terms of xx and yy

  1. Multiply both sides by ss.

    xs=y+5x s = y + 5

  2. Isolate ss.

    s=y+5xs = \frac{y + 5}{x}

Part 5· 2 steps

Solve dxdt=dydt+10\frac{dx}{dt} = \frac{dy}{dt} + 10 for y′=dydty' = \frac{dy}{dt} (speeds in m/s)

  1. Subtract dydt\frac{dy}{dt} from both sides.

    dxdt−dydt=10\frac{dx}{dt} - \frac{dy}{dt} = 10

  2. Isolate y′y'.

    y′=dxdt−10y' = \frac{dx}{dt} - 10

Part 6· 2 steps

Measures of central tendency for data (value, frequency): (2, 6), (3, 10)

  1. Mode is the value with highest frequency.

    Mode = 3 (frequency 10 > 6)

  2. Mean xˉ=∑(x⋅f)∑f\bar{x} = \frac{\sum (x \cdot f)}{\sum f}.

    ∑(x⋅f)=2⋅6+3⋅10=12+30=42\sum (x \cdot f) = 2 \cdot 6 + 3 \cdot 10 = 12 + 30 = 42 ∑f=6+10=16\sum f = 6 + 10 = 16 xˉ=4216=218=2.625\bar{x} = \frac{42}{16} = \frac{21}{8} = 2.625

    Mean = 2.625

    x \approx 1.82 cm
    y \approx 5.32 cm
    c \approx 4.22 cm
    \angle A \approx 14.1^\circ
    \angle B \approx 155.9^\circ
    s = \frac{y+5}{x}
    y' = \frac{dx}{dt} - 10
    mode = 3
    mean = 2.625

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